A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.
Given that:
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}} $$
To prove:
$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = \theta $$
My work so far;
Proof:
$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$
I'm not sure about the above, if there's an easy explanation or method to go through would be great as questions like these are giving me a bit of trouble.
Best Answer
Hint:$$\zeta=\frac{2(\sin2\theta)(\sin2\theta+i\cos2\theta)}{2(\cos\theta)(\sin\theta+i\cos\theta)}=\frac{2\sin 2\theta\cdot ie^{-2i\theta}}{2\cos\theta\cdot ie^{-i\theta}}=2\sin\theta e^{-i\theta}.$$